The given situation can be represented as shown in the following figure

Where, $AO =$ Incident path of the ball

$OB =$ Path followed by the ball after deflection

$\angle AOB =$ Angle between the incident and deflected paths of the ball $=45^{\circ}$

$\angle AOP =\angle BOP =22.5^{\circ}=\theta$

Initial and final velocities of the ball $=v$

Horizontal component of the initial velocity $=v \cos \theta$ along $RO$

Vertical component of the initial velocity $=v \sin \theta$ along $PO$

Horizontal component of the final velocity $=v \cos \theta$ along $OS$

Vertical component of the final velocity $=v \sin \theta$ along $OP$

The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.

$\therefore$ Impulse imparted to the ball $=$ Change in the linear momentum of the ball $=m v \cos \theta-(-m v \cos \theta)$ $=2 m v \cos \theta$

Mass of the ball, $m=0.15\, kg$

Velocity of the ball, $v=54 \,km / h =15 \,m / s$

$\therefore \text { Impulse }=2 \times 0.15 \times 15 \cos 22.5^{\circ}=4.16\, kg\, m / s$